Question:
Two identical short bar magnets, each having a magnetic moment of 10 \( \text{Am}^{2} \) are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane. If the distance between their centres is 0.2 m, the resultant magnetic induction at a point midway between them is: (\( \mu_{\circ}=4\pi\times10^{-7}~\text{Hm}^{-1} \))
Two identical short bar magnets, each having a magnetic moment of 10 \( \text{Am}^{2} \) are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane. If the distance between their centres is 0.2 m, the resultant magnetic induction at a point midway between them is: (\( \mu_{\circ}=4\pi\times10^{-7}~\text{Hm}^{-1} \))
Show Hint
For perpendicular magnetic configurations with a shared distance, the resultant combination always simplifies to a factor of \( \sqrt{2^2 + 1^2} = \sqrt{5} \) times the base equatorial field value, saving you from repeating long scientific calculations.
Updated On: Jun 8, 2026
- \( \sqrt{2}\times10^{-7}~\text{T} \)
- \( \sqrt{5}\times10^{-7}~\text{T} \)
- \( \sqrt{2}\times10^{-3}~\text{T} \)
- \( \sqrt{5}\times10^{-3}~\text{T} \)
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The Correct Option is D
Solution and Explanation
Concept:
The distance from the center of each magnet to the midpoint is \( d = \frac{0.2}{2} = 0.1~\text{m} \).
For a short bar magnet with magnetic moment \( M \):
• The magnetic field at an axial point is \( B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{d^3} \)
• The magnetic field at an equatorial point is \( B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{d^3} \)
Since the magnets are perpendicular, their individual field vectors at the midpoint are mutually perpendicular, and the net field is \( B_{\text{net}} = \sqrt{B_{\text{axial}}^2 + B_{\text{equatorial}}^2} \).
Step 1: Calculating the base field constant factor.
Let \( B_0 = \frac{\mu_0}{4\pi} \frac{M}{d^3} \). Using \( \frac{\mu_0}{4\pi} = 10^{-7} \), \( M = 10\,\text{Am}^2 \), and \( d = 0.1\,\text{m} \): \[ B_0 = 10^{-7} \times \frac{10}{(0.1)^3} = 10^{-7} \times \frac{10}{10^{-3}} = 10^{-7} \times 10^4 = 10^{-3}~\text{T} \]
Step 2: Expressing fields and finding the net resultant vector.
The two individual field magnitudes are:
• \( B_1 = B_{\text{axial}} = 2B_0 \)
• \( B_2 = B_{\text{equatorial}} = B_0 \)
Now, calculate the net combined magnetic field: \[ B_{\text{net}} = \sqrt{(2B_0)^2 + B_0^2} = \sqrt{4B_0^2 + B_0^2} = \sqrt{5B_0^2} = \sqrt{5}B_0 \] Substitute \( B_0 = 10^{-3}~\text{T} \): \[ B_{\text{net}} = \sqrt{5}\times10^{-3}~\text{T} \] This matches option (D) perfectly.
• The magnetic field at an axial point is \( B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{d^3} \)
• The magnetic field at an equatorial point is \( B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{d^3} \)
Since the magnets are perpendicular, their individual field vectors at the midpoint are mutually perpendicular, and the net field is \( B_{\text{net}} = \sqrt{B_{\text{axial}}^2 + B_{\text{equatorial}}^2} \).
Step 1: Calculating the base field constant factor.
Let \( B_0 = \frac{\mu_0}{4\pi} \frac{M}{d^3} \). Using \( \frac{\mu_0}{4\pi} = 10^{-7} \), \( M = 10\,\text{Am}^2 \), and \( d = 0.1\,\text{m} \): \[ B_0 = 10^{-7} \times \frac{10}{(0.1)^3} = 10^{-7} \times \frac{10}{10^{-3}} = 10^{-7} \times 10^4 = 10^{-3}~\text{T} \]
Step 2: Expressing fields and finding the net resultant vector.
The two individual field magnitudes are:
• \( B_1 = B_{\text{axial}} = 2B_0 \)
• \( B_2 = B_{\text{equatorial}} = B_0 \)
Now, calculate the net combined magnetic field: \[ B_{\text{net}} = \sqrt{(2B_0)^2 + B_0^2} = \sqrt{4B_0^2 + B_0^2} = \sqrt{5B_0^2} = \sqrt{5}B_0 \] Substitute \( B_0 = 10^{-3}~\text{T} \): \[ B_{\text{net}} = \sqrt{5}\times10^{-3}~\text{T} \] This matches option (D) perfectly.
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